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a) 4/3 - x = 3/5 + 1/2
=> 4/3 - x= 0,8
=> x = 4/3 + 0/8
=> x = 5/8
TL:
a)
\(\frac{14\left(x-3\right)}{21}+\frac{7\left(x-5\right)}{21}=\frac{13x+4}{21}\)
\(\frac{14x-42+7x-35}{21}=\frac{13x+4}{21}\)
21x-77=13x+4
21x-13x=77+4
8x=81
x=\(\frac{81}{8}\)
2 câu còn lại bn lm cách tương tự
hc tốt
c) Ta có: \(\left\{{}\begin{matrix}\dfrac{x+2}{x+1}+\dfrac{2}{y-2}=6\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{x+1}+\dfrac{10}{y-2}=25\\\dfrac{5}{x+1}-\dfrac{1}{y-2}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{11}{y-2}=22\\\dfrac{1}{x+1}+\dfrac{2}{y-2}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=\dfrac{1}{2}\\\dfrac{1}{x+1}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+1=1\\y-2=\dfrac{1}{2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{5}{2}\end{matrix}\right.\)
\(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\\ \left(x+\frac{1}{5}\right)^2=\frac{9}{25}\\ \left|\left(x+\frac{1}{5}\right)\right|=\frac{3}{5}\)
TH1: \(x=\frac{3}{5}-\frac{1}{5}\\ x=\frac{2}{5}\)
TH2: \(\left|\left(x+\frac{1}{5}\right)\right|=-\frac{3}{5}\\ x=-\frac{3}{5}-\frac{1}{5}\\ x=-\frac{4}{5}\)
\(a,\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Rightarrow\left(x+\frac{1}{5}\right)^2=\left(\frac{3}{5}\right)^2\)
\(\Rightarrow x+\frac{1}{5}=\frac{3}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
\(b,-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}+\frac{24}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Rightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Rightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Rightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Rightarrow3x=\frac{1}{9}\)
\(\Rightarrow x=\frac{1}{27}\)
\(c,\left(x+\frac{1}{2}\right)\left(\frac{2}{3}-2x\right)=0\)
\(\Rightarrow\) \(\left[\begin{array}{nghiempt}x+\frac{1}{2}=0\\\frac{2}{3}-2x=0\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\2x=\frac{2}{3}\end{array}\right.\) \(\Rightarrow\) \(\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=\frac{1}{3}\end{array}\right.\)
a) \(\left(x+\frac{1}{3}\right)^3=\frac{-8}{27}\)
\(\left(x+\frac{1}{3}\right)^3=\left(\frac{-2}{3}\right)^3\)
\(x+\frac{1}{3}=\frac{-2}{3}\)
\(x=-1\)
b) \(\left(\frac{1}{3}x+\frac{4}{3}\right)^2=\frac{25}{9}\)
\(\left(\frac{1}{3}x+\frac{4}{3}\right)^2=\left(\frac{5}{3}\right)^2\)
\(\frac{1}{3}x+\frac{4}{3}=\frac{5}{3}\)
\(\frac{1}{3}x=\frac{1}{3}\)
\(x=1\)
c) \(2^x+2^{x+1}=24\)
\(2^x+2^x.2=24\)
\(2^x.\left(1+2\right)=24\)
\(2^x.3=24\)
\(2^x=8\)
\(2^x=2^3\)
\(x=3\)
a, (x+1/3)^3 = -8/27
=>(x+1/3)^3 = (-2/3)^3
=>x+1/3 = -2/3
=>x = -1
b, (1/3x+4/3)^2 = 25/9
=>(1/3x+4/3)^2 = (5/3)^2
=>(1/3x+4/3) = 5/3
=>1/3x = 1/3
=> x = 1
c, 2^x + 2^x+1 = 24
=>2^x + 2^x . 2 = 24
=>2^x.(1+2) = 24
=>2^x . 3 = 24
=>2^x =8
=>2^x = 2^3
=> x = 3